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Old 10-19-2007, 12:58 AM   #1
CrispyGriz
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So i am in Calculus 113 in university right now and am having huge problems figuring out derivatives. So can one of you math whizzes out there give me a hand. i need to find the second derivative of y for the equation

y=9tan(x/3)

So help away
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Old 10-19-2007, 01:25 AM   #2
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This just comes down to being familiar with the chain rule and knowing the derivatives of tan(x) and sec(x).

y' = 9 sec^2(x/3) (1/3) = 3 sec^2(x/3) = 3 [ sec(x/3) ]^2
y'' = 3 * [ 2 sec(x/3) ] sec(x/3)tan(x/3) * (1/3)
y'' = 2 sec^2(x/3)tan(x/3)

I'm fairly sure that's right.. I'm in Calculus III right now, and believe it or not, the higher level of calc you go into the simpler the derivatives/integrals get.. so it's been awhile.

Basically the chain rule is working from the outside in, multiplying the derivative of each "part".. So you have 9 tan(x/3).. Let u=x/3 if it helps.. so 9tan(u) is one "part". the derivative of 9tan(u) is 9sec^2(u).. now do the next "part", which is "u", which is x/3. the derivative of x/3 is 1/3, so you multiply 9sec^2(u) by 1/3. Get it?

Sorry, I just realized I'm not very good at explaining this.. but i'll try again.

To find the second derivative (y''), you just take the derivative of the derivative. So it's a little bit trickier as it has 3 parts this time.

the first part is.. 3 [blah]^2.. let blah= sec(x/3)
so treat blah, just like x.. and you know the derivative of 3x^2 is 6x right? But instead, we have 6(blah). now we have to differentiate blah.. which is sec(x/3).. the derivative of sec(x/3) is sec(x/3)tan(x/3)(1/3). Kinda the same as when you found the derivative.

I know it's not very clear, but I wish you knew your current understand of calc so i knew where to put more emphasis.

*there's nothing special to know about the derivative of tan(x) and sec(x).. you just have to memorize them. there's usually a list of ones to memorize on the inside of cover of most textbooks.

Edit: there's also sort of a trick to this problem. which is you need to know that sec^2(x) = [sec(x)]^2. This is true with all trig functions. sin^2(x) = [sin(x)]^2, etc.

Last edited by BananaPancakes; 10-19-2007 at 01:29 AM.
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Old 10-19-2007, 01:27 AM   #3
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^I'd give you a red square for that.
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Old 10-19-2007, 01:33 AM   #4
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BananaPancakes is right. I took that same course you are in 6 years ago and the main understanding you need to have is the chain rule in order to solve problems like this. As was mentioned, work from outside in and take the derivative of each term and multiply it. The first portion (3 lines) BananaPancakes explains does this and he has the correct methodology.
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Old 10-19-2007, 01:44 AM   #5
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Thanks for the help, i know where to direct my problems from now on

This calc course has been rough on me, was out of high school 3 years before going back to school and forgot almost all of grade 12 math, so Ive been trying to grasp both at the same time and got lost completely on derivatives
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Old 10-19-2007, 01:58 AM   #6
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Just for confirmation, even though i know i will look like an idiot, but regardless, if i take the derivative of x, would i get 1?
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Old 10-19-2007, 02:08 AM   #7
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Yup, the derivative of x is 1.

if you need some explanation, it's the mini-chain rule thing again..

x=x^1, derivative of x= 1x^(1-1) = 1x^(0) = 1

Not a problem with the help though. I plan on being a math teacher/prof if this engineering thing doesn't work out. Hopefully I'm getting a little better at teaching.

Calculus is a really hard concept to grasp at first, but once you get it, it becomes very interesting, imo.
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Old 10-19-2007, 02:11 AM   #8
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now that im sure on that, everything actually makes sense. The explanation somehow made a lot of sense.

10 points for you
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Old 10-19-2007, 07:53 AM   #9
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Is it kind of sad I couldn't do this problem?

I think it's really sad. I've done all the calculus courses UVic offers...and I can't do this question. Chain Rule? Ugh. Bad memories of first year.

As for the guy with the answers, calc III kicked my ass. I hope you do better than I did. It's a tough course, especially for a guy that sucks at math like I do.

Although I do know d(x)/dx = 1

And btw, the reasoning for that is not the chain rule, but the power rule. Yay I'm good at math!.....NOT.
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Old 10-19-2007, 08:01 AM   #10
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Bahaha.

So glad I never have to take another math course in my life.
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Old 10-19-2007, 09:01 AM   #11
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Quote:
Originally Posted by OutOfTheCube View Post
Bahaha.

So glad I never have to take another math course in my life.
And i officially hate you now
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Old 10-19-2007, 09:05 AM   #12
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I thought I had escaped this stuff after my first year of university over 6 years ago, but now that I'm prepping for the GMAT, it's coming back into my life again.....BLARGH
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Old 10-19-2007, 09:08 AM   #13
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Quote:
Originally Posted by OutOfTheCube View Post
Bahaha.

So glad I never have to take another math course in my life.
I couldnt agree more!
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Old 10-19-2007, 10:44 AM   #14
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Does anyone else see the cosmic irony of learning calculus from a a guy with the handle "BananaPancakes"?

Its not that hes wrong (believe me, I'm far too lazy to check) I just think its hilarious.

You should cite him for his work on your tests.......I'd kill to see the look on that prof's face.....
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Old 10-19-2007, 11:01 AM   #15
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And I thought this topic was going to be about oil futures. O, university was so long ago....
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Old 10-19-2007, 11:05 AM   #16
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Wow, I just had a nightmare flashback to SAIT.

:::wanders off muttering in fear of calculus:::
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Old 10-19-2007, 12:30 PM   #17
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Wow! I thought this might be an investing question I could help with. But no.

I'll leave now. My head is spinning. Good job on the answer. Wish I understood it.
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Old 10-19-2007, 12:47 PM   #18
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Whosh....

Count me as another happy soul that is glad he doesn't have to take ANY math courses again.
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Old 10-19-2007, 02:40 PM   #19
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I can spell cat
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Old 10-19-2007, 02:45 PM   #20
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I can spell cat
But can you spell dog??


*shudder* Calculus....I had 70% going into the final first year, and I failed the final...had to retake it
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